Here's the question:


  • $\overset{\Delta}{ABC}$ is a triangle.
  • $D$ is a point on $[BC]$.
  • $|BD|=4$.
  • $|AD|=|CD|$.
  • $\text m(\widehat{CBA})=\alpha=30^\circ$.
  • $\text m(\widehat{ACB})=\beta=20^\circ$.
  • What is $\color{magenta}{|AC|}=\color{magenta}x$?

Geometric approaches gave me nothing. Letting $|AD|=|CD|=a$, $|AB|=b$ and applying law of cosines to all three triangles yields;

$$ \begin{align} x&=2a\cos 20^\circ\\ 16&=a^2+b^2+2ab\cos 70^\circ\\ (a+4)^2&=b^2+x^2+2bx\cos 50^\circ \end{align} $$

respectively, from $\overset{\Delta}{ACD}$, $\overset{\Delta}{ABD}$ and $\overset{\Delta}{ABC}$. Directly isolating $x$ from these equations gives an expression involves trigonometric function(s), and i cannot find a way to annihilate trigonometric functions in these equations to reach the answer which is $\color{magenta}x=\color{magenta}4$.


4 Answers 4


Here is an elegant purely geometric solution. The diagram shows the 18-gon that I used to find the solution, but the solution itself does not need the 18-gon.

Let $E$ be the image of $B$ under the reflection that maps $A$ to $D$.

Then $\angle EAC = \angle EAD + \angle DAC = \angle ADB + \angle DAC = 40^\circ + 20^\circ = 60^\circ$.

Also $\angle DEA = \angle DBA = 30^\circ$.

Thus $DE$ bisects $AC$ because $DE \perp AC$ and $\triangle ACD$ is isosceles.

Thus $\triangle EAC$ is equilateral and hence $\overline{AC} = \overline{AE} = \overline{BD}$.

  • $\begingroup$ This regular polygon method shows its effectiveness again. However i need some clarification on some point: How did you found $\text m(\widehat{EAD})=40^\circ$ and $\text m(\widehat{DEA})=30^\circ$ without using eighteengon? $\endgroup$
    – Alistair
    Dec 1, 2014 at 23:46
  • $\begingroup$ @Alistair: Both are by the reflection. I can add one more intermediate step to make it all clear. $\endgroup$
    – user21820
    Dec 2, 2014 at 10:59

enter image description hereDrop a perpendicular from $D$ to $M$ on $AC$. By congruence of the resultant $\bigtriangleup AMD$ and $\bigtriangleup MDC\,(*)$ we observe that $|AM| = x/2$ and so $|AD| = (x/2)\sec20^{\circ}$.

Now we note that $A\hat{D}B = 20\times2 = 40^{\circ}$ since the exterior angle of a triangle equals the sum of $2$ opposite interior angles.

So by the angle sum of $\bigtriangleup ABD,\,B\hat{A}D = 180-30-40 = 110^{\circ}$. Then, applying the sine rule: $$\frac{AD}{\sin A\hat{B}D} = \frac{4}{\sin B\hat{A}D}$$ we have $$ \frac{(x/2)\sec20^{\circ}}{\sin30^{\circ}} = \frac{4}{\sin 110^{\circ}}$$ which by rearrangement, yields $|AC|$:$$x = 8\sin 30^{\circ} \cos20^{\circ} \mathrm{cosec}\,110^{\circ} = 4$$

$(*)$ The congruence can be seen from the facts that:

  • $|AD| = |DC|$
  • $|MD|$ is a common length
  • $A\hat{M}D = D\hat{M}C$ since $DM\perp AC$

satisfying the $RHS$ test for right triangles.

  • $\begingroup$ Thanks for clear answer. I believe this is the most direct way to reach $x$ without using trigonometric identities. I had a feeling that this was an easy question and i'm missing something obvious like law of sines exterminated by my brain somehow. $\endgroup$
    – Alistair
    Dec 1, 2014 at 23:43

First calculate some angles:

$ \angle DAC = \angle DCA = 20^o $ (isosceles triangle theorem https://en.wikipedia.org/wiki/Isosceles_triangle )

$ \angle CAD = 180^o -20^o -20^o = 140^o $ (triangle is $180^o$)

$ \angle CAB = 180^o -30^o -20^o = 130^o $ (triangle is $180^o$)

$ \angle BAD = \angle BAC- \angle DAC = 130^o-20^o = 110^o $

and the rest is all the law of sinus ( https://en.wikipedia.org/wiki/Law_of_sines ):

$ \frac{AD}{\sin (\angle ABD)} = \frac{BD} {\sin(\angle BAD)} $

$ \frac{AD}{\sin (\angle ACD)} = \frac{AC} {\sin(\angle ADC)} $

done :)

  • 4
    $\begingroup$ The sinus is in your nose and does not have any rule. =) $\endgroup$
    – user21820
    Dec 1, 2014 at 10:48
  • 1
    $\begingroup$ Feeling a nos-e-ous after cracking up at this. $\endgroup$
    – Hernandez
    Dec 1, 2014 at 10:53
  • $\begingroup$ For some mysterious reason sine rule completely left my mind. This was an easy question as you implicitly imply. $\endgroup$
    – Alistair
    Dec 1, 2014 at 23:26

Construct isoceles $\triangle BDF$ with common angle $20^\circ$ and isoceles $\triangle BDG$ with common angle $30^\circ$. Now $\angle BGF=\angle FGD=\angle DGA=60^\circ$, and $\angle GDF=\angle ADG=10^\circ$. So $\triangle GDF\cong \triangle GDA$ and so $\overline{DF}=\overline{DA}$. Since $\triangle ADC\sim\triangle DFB$, we have the desired result.


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